0

Let $\mathcal{S_3}$ be the collection of $3\times 3$ skew-symmetric matrices with integer entries. There is no non-zero $D\in \mathcal{S_3}$ that satisfies the following equation $$[D,[D,[D,N]]]=0$$ for all $N\in \mathcal{S_3}$ where $[\cdot,\cdot]$ is the usual commutator of matrices ($[A,B]=AB-BA$).

How can I prove or disprove this for all $n$, i.e., show or disprove that there is no $D\in \mathcal{S_n}\setminus {0}$ such that $[D,[D,[D,N]]]=0$ for any $N\in \mathcal{S_n}$?

I've attempted to prove it by considering the generators of $\mathcal{S_n}$ and then computing its commutator with one another. I then expressed $D$ as the linear combination of those generators. It is very tedious (even if I let $N$ to be one of the generators) and I did not get any result. I am now looking for other ways to prove or disprove it. Does $\mathcal{S_n}$ have properties you that might be useful in this problem?

Elliot
  • 1
  • 1
  • 1
    Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be closed. To prevent that, please [edit] the question. This will help you recognize and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. – José Carlos Santos Aug 03 '22 at 04:29
  • The equality implies $ad(D)^3=0$ in the Lie algebra $\mathfrak{so}(n)$. So the linear operator $ad(D)$, defined by $ad(D)(N)=[D,N]=DN-ND$, consisting of a skew-symmetric matrix, is nilpotent. It follows that $ad(D)=0$, because the only skew-symmetric nilpotent matrix is the zero matrix. It follows that $D=0$, a contradiction. – Dietrich Burde Aug 03 '22 at 11:57

0 Answers0